tradinggenius wrote:
A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?
1) 7/40
2) 7/20
3) 49/100
4) 21/40
5) 7/10
Probability = # of favorable outcomes / total # of outcomes;
P=\frac{C^2_7*C^1_3}{C^3_{10}}=\frac{21}{40}, where
C^2_7 is # of ways to choose 2 different green apples out of 7,
C^1_3 is # of ways to choose 1 red apple out of 3, and
C^3_{10} is total # of ways to choose 3 different apples out of total 10 apples.
Answer: D.
Or by probability approach:
P(GGR)=\frac{3!}{2!}*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}=\frac{21}{40}, we are multiplying by 3!/2! as the case of GGR can occur in 3 ways: GGR, GRG, RGG - # of permutation of 3 letters out of which 2 are identical.
Answer: D.
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92% (01:44) correct
